# Killing vecotrs of Schwarzschild metric

1. Apr 1, 2014

### Vrbic

Hello,
I have a question, whether is possible to looking for Killing Vectors (KV) in this way (I know about general solution):
From Schwarzschild metric I can see two KV $\frac{\partial}{\partial t}$ and $\frac{\partial}{\partial\phi}$. Then I see that other trivial KV arent there. Metric in dt and dr is independant on $\theta, \phi$ so I suppose I can "split" metric and looking for KV just in spherical part $d\theta^2+\sin^2{\theta}d\phi^2$.
Can I suppose transformation this metric to the form: $d\alpha^2+d\beta^2$ and claim the $\frac{\partial}{\partial\alpha}, \frac{\partial}{\partial\beta}$ are KV?

2. Apr 1, 2014

### WannabeNewton

No. You won't be able to find such a transformation.

Those aren't the only trivial Killing fields by the way. The generators of rotations, that is, the basis of $so(3)$, are also easily seen to be Killing fields of the Schwarzschild metric and they arise from the spherical symmetry.

3. Apr 1, 2014

### Vrbic

Do I need know all transformation? I thought that differencial form of it could be enough. If I suppose $\alpha=\alpha(\theta,\phi), \beta=\beta(\theta,\phi)$, than I can write $d\alpha=\alpha_{\theta}+\alpha_{\phi}$, $d\beta=\beta_{\theta}+\beta_{\phi}$, where subscript means differentation. When I put it back to the $ds^2=d\alpha^2+d\beta^2$ I get 3 eq.:
$1=\alpha_{\theta}^2+\beta_{\theta}^2$
$0=\alpha_{\theta}\alpha_{\phi}+\beta_{\theta}\beta_{\phi}$
$\sin^2{\theta}=\alpha_{\phi}^2+\beta_{\phi}^2$.
From them (if from first guess $\alpha_{\theta}=\cos{\phi}$, $\beta_{\theta}=\sin{\phi}$) I get $d\alpha=\cos{\phi}d\theta-\sin{\theta}\sin{\phi}d\phi$, $d\beta=\sin{\theta}d\theta+\sin{\theta}\cos{\phi}d\phi$.
Multiplying by $g^{\mu\nu}$ I hope can get $\frac{\partial}{\partial\alpha}$, $\frac{\partial}{\partial\beta}$. What is wrong?

4. Apr 1, 2014

### Bill_K

If you could put the metric in that form, it would be FLAT. It's not flat, it's a sphere.

5. Apr 1, 2014

### WannabeNewton

This clearly doesn't reduce to the metric on the 2-sphere: simple substitution yields $d\alpha^2 + d\beta^2 = d\theta ^2 + \sin^2 \theta d\phi^2 + (\sin \theta \cos\phi - \sin 2\phi )\sin\theta d\theta d\phi$.

You simply won't be able to put the 2-sphere metric in the form $ds^2 = d\alpha^2 + d\beta^2$ and Bill has elucidated above why this is impossible.